Question

Find the distance between the lines 7x-8y+12=0 and 7x-8y-8=0

Answer 1

$[/tex] answer is [tex]20 \frac{1}{1 \sqrt[]{113} }$

Answer 2

Answer:

The distance between two lines is 0.17.

Step-by-step explanation:

Given : The lines 7x-8y+12=0 and 7x-8y-8=0  

To find : The distance between the lines?

Solution :

If the lines are in form :

ax+by=c and ax+by=d

Then the lines are parallel and distance is given as

$d=\frac{|d-c|}{a^2+b^2}$

On Comparing with given lines,

a=7 , b=-8, c=-12 and d=8

Substitute in the formula,

$d=\frac{|8-(-12)|}{7^2+(-8)^2}$

$d=\frac{|20|}{49+64}$

$d=\frac{20}{113}$

$d=0.17$

Therefore, The distance between two lines is 0.17.